leetcode NO.199二叉树的右视图

层序遍历二叉树后，取每一层节点的最后一个即可

class Solution {
public:
    vector<int> rightSideView(TreeNode* root) {
        vector< vector<int>> temp01;
        vector<int> result;
        queue<TreeNode *> w;
        if(root)
        w.push(root);
        while(! w.empty())
        {
            vector<int> temp02;
            int size = w.size();
            TreeNode * node;
            for(int i = 0;i < size;i ++)
            {
                node = w.front();
                if(node->left)
                w.push(node->left);
                if(node->right)
                w.push(node->right);
                temp02.push_back(node->val);
                w.pop();
            }
            temp01.push_back(temp02);
        }
        for(int i = 0;i < temp01.size();i ++)
            result.push_back(temp01[i][temp01[i].size() - 1]);
        return result;
    }
};

改进：直接在循环中判断是否为每一层的最后一个，不用再另外开辟两个数组完整储存整个二叉树

vector<int> result;
        queue<TreeNode *> w;
        if(root)
        w.push(root);
        while(! w.empty())
        {
            int size = w.size();
            TreeNode * node;
            for(int i = 0;i < size;i ++)
            {
                node = w.front();
                if(node->left)
                w.push(node->left);
                if(node->right)
                w.push(node->right);
                if(i == size - 1)
                result.push_back(node->val);
                w.pop();
            }
        }
        return result;